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The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at a certain instant, P = 8.0 atm and is increasing at a rate of 0.14 atm/min and V = 13 L and is decreasing at a rate of 0.16 L/min. Find the rate of change of T with respect to time at that instant if n = 10 mol. (Round your answer to four decimal places.)
[tex]\frac{dT}{dt}[/tex] = _____ K/min

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debrubs97

Answer:

The temperature of the gas is increasing 0.6577 kelvin/min

Step-by-step explanation:

We have to find the rate of change of T with respect to time at that instant so we should start by arranging the equation[tex]T=\frac{PV}{nR}[/tex]

Now we can derivate T to obtain T'

[tex]T'=\frac{(PV)'(nR)-(PV)(nR)'}{((nR)^{2} }[/tex]

Now we should solve the derivate of (PV) and (nR) by using the product rule

[tex]T'=\frac{(P'V+PV')(nR)-(PV)(n'R+nR')}{(nR)^{2} }[/tex]

We know that

P=8atm

V=13L

P'=0.14

V'=-0.16

n=10

R=0.0821

n'=0

R'=0

We substitute into the equation

[tex]T'=\frac{(0.14*13+8*-0.16)(10*0.0821)-(8*13)(0*0.0821+10*0)}{(10*0.0821)^{2} }[/tex]

[tex]T= 0.6577[/tex]  

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